Kienastsrx

2022-03-16

What is the instantaneous rate of change of $f\left(x\right)=\frac{x}{-x+7}$ at x=0?

nida0694ii5

The instantaneous rate of change is the first derivative calculated at
x=0
Hence
${f}^{\prime }\left(x\right)=\frac{7}{{\left(7-x\right)}^{2}}$
so ${f}^{\prime }\left(0\right)=\frac{7}{{7}^{2}}=\frac{1}{7}$

Talaminiu2d

Apply the quotient rule, which is:
$\frac{d}{dx}\frac{f\left(x\right)}{g\left(x\right)}=\frac{-f\left(x\right)\frac{d}{dx}g\left(x\right)+g\left(x\right)\frac{d}{dx}f\left(x\right)}{{g}^{2}\left(x\right)}$
f(x)=x and g(x)=7-x
To find $\frac{d}{dx}f\left(x\right):$
Apply the power rule: x goes to 1
To find $\frac{d}{dx}g\left(x\right):$
Differentiate 7-x term by term:
The derivative of the constant 7 is zero.
The derivative of a constant times a function is the constant times the derivative of the function.
Apply the power rule: x goes to 1
So, the result is: -1
The result is: -1
Now plug in to the quotient rule:
$\frac{7}{{\left(7-x\right)}^{2}}$
Now simplify:
$\frac{7}{{\left(x-7\right)}^{2}}$
If we putting the value of x, we get:
0

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